The triangles are congruent by the SAS congruence theorem. By extension, we can also say they are similar using a closely related theorem. Any time there are congruent triangles, they are automatically similar.
This is why the answer is choice C) Similar - SAS
Note how
SB = ET ... which forms the first "S" in "SAS"
angle ESB = angle SET ... which forms the "A" in "SAS"
SE = SE ... which forms the second "S" in "SAS"
this is why SAS works
Answer:

Step-by-step explanation:
Given the following mathematical function;

<u>When x = 1</u>


<u>When x = 2</u>



<u>When x = 3</u>



<u>When x = 4</u>



<u>When x = 5</u>



We are given that there
will be (1/2) a litre after the first pouring, so considering two successive
pourings (n and (n+1)) with 1/2 litre in each before the nth pouring occurs:
1/2 × (1/n) = 1/(2n)
1/2 - 1/(2n) = (n-1)/2n
1/2 + 1/(2n) = (n+1)/2n
(n-1)/2n and (n+1)/2n in
each urn after the nth pouring
Then now consider the
(n+1)th pouring
(n+1)/2n × 1/(n+1) =
1/(2n)
(n+1)/(2n) - 1/(2n) =
n/(2n) = 1/2
Therefore this means that after
an odd number of pouring, there will be 1/2 a litre in each urn
Since 1997 is an odd
number, then there will be 1/2 a litre of water in each urn.
Answer:
<span>1/2</span>
Answer:
No
Step-by-step explanation:
It can't because no number goes into 24 and 7, and 7 does not go its 24
The percent error is the absolute difference of the actual and measured values, divided by the actual value, multiplied by 100. In here, the actual value is 44.5 lb, while the measured value is 46 pounds.
Percent error = |44.5 - 46|/44.5 * 100 = <em>3.37%</em>